Question

question 4 1 point
find all the solutions, including imaginary solutions if they exist, of the following equation.
$x^{4}+20x^{2}+64 = 0$
list your answers separated by commas, not using a ± sign.
for example, if you found that $x=pm1$ or $x = pm i$, you would enter 1, -1, i, -i.
provide your answer below:

Explanation:

Step1: Let \(u = x^{2}\)

The equation \(x^{4}+20x^{2}+64 = 0\) becomes \(u^{2}+20u + 64=0\).

Step2: Factor the quadratic equation

We factor \(u^{2}+20u + 64\) as \((u + 4)(u+16)=0\).

Step3: Solve for \(u\)

Using the zero - product property, if \((u + 4)(u + 16)=0\), then \(u+4=0\) or \(u + 16=0\). So \(u=-4\) or \(u=-16\).

Step4: Substitute back \(u = x^{2}\)

When \(u=-4\), we have \(x^{2}=-4\), so \(x=\pm2i\). When \(u=-16\), we have \(x^{2}=-16\), so \(x=\pm4i\).

Answer:

\(2i,-2i,4i,-4i\)